Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n;
int a[10005];
int b[10005];
pair<int, int> ans1[1000005], ans2[1000005];
int s1, s2;
void gauss(int *a, pair<int, int> *ans, int &s) {
  int w = 1;
  for (int i = 0; i <= 31; i++) {
    int noww = -1;
    for (int j = w; j <= n; j++) {
      if (a[j] & (1 << i)) {
        noww = j;
        break;
      }
    }
    if (noww == -1) continue;
    if (noww != w) {
      swap(a[w], a[noww]);
      s++;
      ans[s] = make_pair(w, noww);
      s++;
      ans[s] = make_pair(noww, w);
      s++;
      ans[s] = make_pair(w, noww);
    }
    for (int j = 1; j <= n; j++) {
      if (j == w) continue;
      if (a[j] & (1 << i)) {
        a[j] = a[j] ^ a[w];
        s++;
        ans[s] = make_pair(j, w);
      }
    }
    w++;
  }
}
void solve() {
  for (int i = 1; i <= 32; i++) {
    int w = -1;
    for (int j = 0; j <= 31; j++) {
      if (a[i] & (1 << j)) {
        w = j;
        break;
      }
    }
    if (w == -1) continue;
    for (int j = 1; j <= 32; j++) {
      if ((a[j] & (1 << w)) ^ (b[j] & (1 << w))) {
        a[j] = a[j] ^ a[i];
        s1++;
        ans1[s1] = make_pair(j, i);
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  gauss(a, ans1, s1);
  gauss(b, ans2, s2);
  solve();
  for (int i = 1; i <= n; i++) {
    if (a[i] != b[i]) {
      puts("-1");
      return 0;
    }
  }
  printf("%d\n", s1 + s2);
  for (int i = 1; i <= s1; i++)
    printf("%d %d\n", ans1[i].first, ans1[i].second);
  for (int i = s2; i >= 1; i--)
    printf("%d %d\n", ans2[i].first, ans2[i].second);
  return 0;
}